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%I #18 Aug 30 2017 12:15:32
%S 1,2,3,6,7,23,19,66,95,255,187,1059,631,3227,5243,11426,7711,51887,
%T 27595,184911,232887,606627,364723,2807935,2405183,8671943,10368079,
%U 36873651,18512791,167268639,69273667,496472226,551130063,1856103039
%N a(n) = A159068(n)/n.
%H Seiichi Manyama, <a href="/A159069/b159069.txt">Table of n, a(n) for n = 1..1000</a>
%e Row 6 of Pascal's triangle is 1,6,15,20,15,6,1. The greatest common divisors of n and each integer from 1 to 6 are gcd(1,6)=1, gcd(2,6)=2, gcd(3,6)=3, gcd(4,6)=2, gcd(5,6)=1, and gcd(6,6)=6. So a(6) = (1/6)*( 6*1 + 15*2 + 20*3 + 15*2 + 6*1 + 1*6) = 138/6 = 23. Note that each term of the sum in parentheses is a multiple of 6, so 138 is a multiple of 6.
%p A159068 := proc(n) add(binomial(n,k)*gcd(k,n),k=1..n) ; end: A159069 := proc(n) A159068(n)/n ; end: seq(A159069(n),n=1..80) ; # _R. J. Mathar_, Apr 06 2009
%t Table[Sum[Binomial[n, k] GCD[k, n], {k, n}]/n, {n, 34}] (* _Michael De Vlieger_, Aug 29 2017 *)
%o (PARI) a(n) = sum(k=1, n, binomial(n,k) * gcd(k,n))/n; \\ _Michel Marcus_, Aug 30 2017
%Y Cf. A159068.
%K nonn
%O 1,2
%A _Leroy Quet_, Apr 04 2009
%E Extended by _R. J. Mathar_, Apr 06 2009
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